Prove. (1 + tanA tanB)^2 + (tanA - tanB)^2 = sec^2A sec^2B - Vidyadip

Question

Prove.
$(1 + tanA\ tanB)^2 + (tanA - tanB)^2 = sec^2A sec^2B$

✓

Answer

$LHS =(1 + tanA tanB)^2 + (tanA - tanB)^2$
$= 1 + \tan^2A \tan^2B + 2tanA\ tanB+\tan^2A + \tan^2B - 2 tanA tanB$
$= 1 + \tan^2A + \tan^2B + \tan^2A \tan^2B$
$= sec^2A + \tan^2B (1 + \tan^2A)$
$= sec^2A + \tan^2B sec^2A$
$= sec^2A (1 + \tan^2B)$
$= sec^2A sec^2B = RHS$

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